FACTS  AND  FORMULAE  FOR  HCF  AND  LCM  QUESTIONS

 

 

I.Factors and Multiples : If a number 'a' divides another number 'b' exactly, we say that 'a' is a factor of 'b'. In this case, b is called a multiple of a.

 

II.Highest Common Factor (H.C.F) or Greatest Common Measure (G.C.M) or Greatest Common Divisor (G.C.D) : The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly. There are two methods of finding the H.C.F. of a given set of numbers :

1. Factorization Method : Express each one of the given numbers as the product of prime factors.The product of least powers of common prime factors gives H.C.F.

2. Division Method : Suppose we have to find the H.C.F. of two given numbers. Divide the larger number by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is the required H.C.F.

Finding the H.C.F. of more than two numbers : Suppose we have to find the H.C.F. of three numbers. Then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given numbers. Similarly, the H.C.F. of more than three numbers may be obtained. 

 

III.Least Common Multiple (L.C.M) : The least number which is exactly divisible by each one of the given numbers is called their L.C.M.

1. Factorization Method of Finding L.C.M: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.

2. Common Division Method (Short-cut Method) of Finding L.C.M : Arrange the given numbers in a row in any order. Divide by a number which divides exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.

 

IV. Product of two numbers = Product of their H.C.F and L.C.M

 

V. Co-primes : Two numbers are said to be co-primes if their H.C.F. is 1.

 

VI. H.C.F and L.C.M of Fractions :

1. H.C.F = H.C.F. of Numerators / L.C.M of Numerators

2. L.C.M = L.C.M of Numerators / H.C.F of Denominators

 

VII. H.C.F and L.C.M of Decimal Fractions : In given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.

 

VIII. Comparison of Fractions : Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M. as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

Q:

Find the Greatest Number that will devide 43, 91  and 183  so as to leave the same remainder in each case

A) 4 B) 7
C) 9 D) 13
 
Answer & Explanation Answer: A) 4

Explanation:

Required Number = H.C.F  of  (91- 43), (183- 91) and (183-43)

                          = H.C.F of 48, 92, and 140 = 4

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33 12330
Q:

The LCM of the three numbers 16, 28 and 42 is

 

A) 168 B) 2
C) 252 D) 336  
 
Answer & Explanation Answer: D) 336  

Explanation:
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1 11286
Q:

The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is

A) 1677 B) 1683
C) 2523 D) 3363
 
Answer & Explanation Answer: B) 1683

Explanation:

 L.C.M of 5, 6, 7, 8 = 840

 

Therefore, Required Number is of the form 840k+3.

 

Least value of k for which (840k+3) is divisible by 9 is k = 2 

 

Therefore, Required  Number = (840 x 2+3)=1683

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38 11280
Q:

Three numbers are in the ratio of 3:4:5 and their L.C.M is 3600.Their HCF is:

A) 40 B) 60
C) 100 D) 120
 
Answer & Explanation Answer: B) 60

Explanation:

Let the numbers be 3x, 4x, 5x.

 

Then, their L.C.M = 60x.

 

So, 60x=3600 or x=60.

 

Therefore,  The numbers are (3 x 60), (4 x 60), (5 x 60).

 

Hence,required H.C.F=60

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14 10624
Q:

A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in Width. Find the least number of square tiles of equal size required to cover the entire floor of the room.

A) 107 B) 117
C) 127 D) 137
 
Answer & Explanation Answer: B) 117

Explanation:

Let us calculate both the length and width of the room in centimeters.

Length = 6 meters and 24 centimeters = 624 cm

width = 4 meters and 32 centimeters = 432 cm

As we want the least number of square tiles required, it means the length of each square tile should be as large as possible.Further,the length of each square tile should be a factor of both the length and width of the room.

Hence, the length of each square tile will be equal to the HCF of the length and width of the room = HCF of 624 and 432 = 48

Thus, the number of square tiles required = (624 x 432 ) / (48 x 48) = 13 x 9 = 117

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20 10598
Q:

LCM of 252, 308 and 198 is

A) 1443 B) 2124
C) 2545 D) 2772
 
Answer & Explanation Answer: D) 2772

Explanation:

Prime factors of 

252 = 2 x 2 x 3 x 3 x 7; 

308 = 2 x 2 x 7 x 11; 

198 = 2 x 3 x 3 x 11

Therefore LCM of 252, 308 and 198 is 2 x 2 x 3 x 3 x 7 x 11 = 4 x 9 x 77 = 2772.

 

 

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20 10325
Q:

A heap of coconuts is divided into groups of 2, 3 and 5 and each time one coconut is left over. The least number of Coconuts in the heap is  ?

A) 63 B) 31
C) 16 D) 27
 
Answer & Explanation Answer: B) 31

Explanation:

To get the least number of coconuts :
LCM = 30 => 30 + 1 = 31

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33 10161
Q:

If the HCF of 210 and 55 is expressible in the form of 210 x 5 + 55P, then value of P = ?

A) -23 B) 27
C) 16 D) -19
 
Answer & Explanation Answer: D) -19

Explanation:

HCF of 210 and 55 is 5

Now, 210x5 + 55P = 5

=> 1050 + 55P = 5

=> 55P = -1045

=> P = -1045/55

=> P = -19.

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28 10144